Difference between revisions of "2022 AMC 10B Problems/Problem 12"
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==Solution 2 (99% Accurate Guesswork)== | ==Solution 2 (99% Accurate Guesswork)== | ||
− | Let's try the answer choices. We can quickly find that when we roll <math>3</math> dice, either the first and second sum to <math>7</math>, the first and third sum to <math>7</math>, or the second and third sum to <math>7</math>. There are <math>6</math> ways for the first and second dice to sum to <math>7</math>, <math>6</math> ways for the first and third to sum to <math>7</math>, and <math>6</math> ways for the second and third dice to sum to <math>7</math>. However, we overcounted (but not by much) so we can assume that the answer is <math>\boxed {\textbf{(C) }4}</math> | + | Let's try the answer choices. We can quickly find that when we roll <math>3</math> dice, either the first and second sum to <math>7</math>, the first and third sum to <math>7</math>, or the second and third sum to <math>7</math>. There are <math>6</math> ways for the first and second dice to sum to <math>7</math>, <math>6</math> ways for the first and third to sum to <math>7</math>, and <math>6</math> ways for the second and third dice to sum to <math>7</math>. However, we overcounted (but not by much) so we can assume that the answer is <math>\boxed {\textbf{(C) }4}</math>. |
~Arcticturn | ~Arcticturn | ||
− | |||
==Solution 3== | ==Solution 3== | ||
− | We can start by figuring out what the probability is for each die to add up to <math>7</math> if there is only <math>1</math> roll. We can quickly see that the probability is <math>\frac16</math> , as there are <math>6</math> ways to make <math>7</math> from <math>2</math> numbers on a die, and there are a total of <math>36</math> ways to add <math>2</math> numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for rolling a sum of <math>7</math> in <math>n</math> rolls would be <math>\frac16</math><math>n</math>. The smallest number that satisfies this is <math>\boxed {\textbf{(C) }4}</math> | + | We can start by figuring out what the probability is for each die to add up to <math>7</math> if there is only <math>1</math> roll. We can quickly see that the probability is <math>\frac16</math>, as there are <math>6</math> ways to make <math>7</math> from <math>2</math> numbers on a die, and there are a total of <math>36</math> ways to add <math>2</math> numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for rolling a sum of <math>7</math> in <math>n</math> rolls would be <math>\frac16</math><math>n</math>. The smallest number that satisfies this is <math>\boxed {\textbf{(C) }4}</math>. |
~mihikamishra | ~mihikamishra | ||
+ | |||
+ | ==Solution 4 (Fakesolve)== | ||
+ | On each roll, there is a <math>\frac 16</math> chance of rolling a sum of <math>7</math>. You would need <math>4</math> of these rolls to get <math>4 \cdot \frac 16,</math> which is larger than <math>\frac 12.</math> Therefore, the answer is <math>\boxed {\textbf{(C) }4}.</math> | ||
+ | |||
+ | ~dbnl | ||
==Video Solution (⚡️Just 4 min⚡️)== | ==Video Solution (⚡️Just 4 min⚡️)== |
Latest revision as of 23:27, 21 October 2024
Contents
[hide]Problem
A pair of fair -sided dice is rolled times. What is the least value of such that the probability that the sum of the numbers face up on a roll equals at least once is greater than ?
Solution 1 (Complement)
Rolling a pair of fair -sided dice, the probability of getting a sum of is Regardless what the first die shows, the second die has exactly one outcome to make the sum We consider the complement: The probability of not getting a sum of is Rolling the pair of dice times, the probability of getting a sum of at least once is
Therefore, we have or Since the least integer satisfying the inequality is
~MRENTHUSIASM
Solution 2 (99% Accurate Guesswork)
Let's try the answer choices. We can quickly find that when we roll dice, either the first and second sum to , the first and third sum to , or the second and third sum to . There are ways for the first and second dice to sum to , ways for the first and third to sum to , and ways for the second and third dice to sum to . However, we overcounted (but not by much) so we can assume that the answer is .
~Arcticturn
Solution 3
We can start by figuring out what the probability is for each die to add up to if there is only roll. We can quickly see that the probability is , as there are ways to make from numbers on a die, and there are a total of ways to add numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for rolling a sum of in rolls would be . The smallest number that satisfies this is .
~mihikamishra
Solution 4 (Fakesolve)
On each roll, there is a chance of rolling a sum of . You would need of these rolls to get which is larger than Therefore, the answer is
~dbnl
Video Solution (⚡️Just 4 min⚡️)
~Education, the Study of Everything
Video Solution by Interstigation
Video Solution by TheBeautyofMath
https://youtu.be/Mi2AxPhnRno?t=207
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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